Banach lemma

I have some problems with understanding proof of Banach lemma: Let's A,
B - non empty sets, f:A->B, g:B->A be a "1-1" functions, then there are
sets A1, B1, A2, B2 A=A1+A2, B=B1+B2 (A1*A2=empty_set, B1*B2=empty_set)
that f(A1)=B1 and g(B2)=A2.

Proof:
a from A is called "extended", when a is in g(B), and g^{-1}(a) is in
f(A). Let's define a*=f^{-1}(g^{-1}(a)) - extension of a. Let's
a_0,a_1,...,a_n be a following sequence: a_0=a, a_i=(a_{i-1})*. Let's
n(a) - number of last term in this sequence (last feasible extension,
when this number is infinite we put n(a)=N) and s(a)=a_{n(a)}. Then
A2={a from A: n(a)=N or (s(a) is in g(B) and g^{-1}(a) isn't in f(A))},
A1=A-A1, B1=f(A1), B2=g(B2) ... Follow-up of this proof is quite easy,
but ...

I can't imagine what is the interpretation of set A2 ... I've been
taking some simple sets like A=B=[0,1] and simple functions f, g like
f(x)=x/2 and g(x)=(1-x/2)/2 and I can't construct A2 ... Why A2 was
define in this way ???

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